Convergence of a 2nd Order Perturbation Theory Sum

[IHateMayonnaise]

Homework Statement

This is a 2nd order perturbation theory question, but I've got it boiled down to the following sum, which I need to determine where (and if) it converges:

$$E_n^{(2)}=\sum_{n\not= m}\left[\frac{Sin^2\left(\frac{n\pi}{4}\right)Sin^2\left(\frac{m\pi}{4}\right)}{n^2-m^2}\right]$$

Homework Equations

The Attempt at a Solution

I am to compare my analytic ("exact") solution to the second order perturbation, which comes out to the equation above multiplied by some constants. Apparently, all I have to do is sum the first few terms and I should be able to see it..but I'm not. To me it seems as though this sum is always equal to zero, since in for the example of the first first and second result:

For $n=1,m=2$
$$E_n^{(2)}=-\frac{1}{6}$$

and also

$n=2,m=1$
$$E_n^{(2)}=\frac{1}{6}$$

such that the first two terms sum to zero. I don't see how this would change for any higher terms. Any thoughts? I can give more detail of the problem if needed, but hopefully this is enough. Thanks

IHateMayonnaise

 

[Mark44]

Here's an idea that might be helpful. The general term in your series satisfies:
$$\frac{-1}{n^2 - m^2}~<~\frac{sin^2(n\pi/4)~sin^2(m\pi/4)}{n^2 - m^2}~<~\frac{1}{n^2 - m^2}$$

Furthermore, you can break the latter expression up by partial fractions decomposition:
$$\frac{1}{n^2 - m^2}~=~\frac{A}{n - m} + \frac{B}{n + m}$$
and solve for A and B.

Hope this helps.

 

[IHateMayonnaise]

Here's an idea that might be helpful. The general term in your series satisfies:
$$\frac{-1}{n^2 - m^2}~<~\frac{sin^2(n\pi/4)~sin^2(m\pi/4)}{n^2 - m^2}~<~\frac{1}{n^2 - m^2}$$

Furthermore, you can break the latter expression up by partial fractions decomposition:
$$\frac{1}{n^2 - m^2}~=~\frac{A}{n - m} + \frac{B}{n + m}$$
and solve for A and B.

Hope this helps.

Gets me started, thanks!